Question

What is the derivative of $3{{x}^{2}}$ using first principal?

Answer 1

We use the formula of $\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$ to find the derivative of $3{{x}^{2}}$. We take a constant common and then use the theorem ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ to simplify the numerator. We put the limit value to find the final solution.

Complete step-by-step solution:
The derivative form of $\dfrac{dy}{dx}$ tends to a definite finite limit when $\Delta x \to 0$, then the limiting value obtained by this can also be found by first order derivatives. We can also apply first order derivative principle to get the differentiated value of $f\left( x \right)=3{{x}^{2}}$.
We know that $\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$. Here $f\left( x \right)=3{{x}^{2}}$. Also, $f\left( x+h \right)=3{{\left( x+h \right)}^{2}}$.
So, $\dfrac{df}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}=\displaystyle \lim_{h \to 0}\dfrac{3{{\left( x+h \right)}^{2}}-3{{x}^{2}}}{h}$.
We first take 3 common from the numerator and get $\displaystyle \lim_{h \to 0}\dfrac{3{{\left( x+h \right)}^{2}}-3{{x}^{2}}}{h}=3\displaystyle \lim_{h \to 0}\dfrac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h}$
We now have to simplify the numerator of the limit. We use the formula ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$.
Therefore, ${{\left( x+h \right)}^{2}}-{{x}^{2}}=\left( x+h+x \right)\left( x+h-x \right)=h\left( 2x+h \right)$.
$3\displaystyle \lim_{h \to 0}\dfrac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h}=3\displaystyle \lim_{h \to 0}\dfrac{h\left( 2x+h \right)}{h}=3\displaystyle \lim_{h \to 0}\left( 2x+h \right)$.
Now we use the limit value and get $3\displaystyle \lim_{h \to 0}\dfrac{{{\left( x+h \right)}^{2}}-{{x}^{2}}}{h}=3\displaystyle \lim_{h \to 0}\left( 2x+h \right)=6x$.
Therefore, derivative of $3{{x}^{2}}$ is $6x$.

Note: Differentiation, the fundamental operations in calculus deals with the rate at which the dependent variable changes with respect to the independent variable. The measurement quantity of its rate of change is known as derivative or differential coefficients. We find the increment of those variables for small changes.
We know the limit value $\displaystyle \lim_{x \to a}\dfrac{{{x}^{n}}-{{a}^{n}}}{x-a}=n{{a}^{n- 1}}$ which gives $\dfrac{df}{dx}=n{{x}^{n-1}}$.
Therefore, $\dfrac{d}{dx}\left( 3{{x}^{2}} \right)=3\times 2\times {{x}^{2-1}}=6x$.
Thus, the derivative of the function $f\left( x \right)=3{{x}^{2}}$ is verified.